let a= sqrt(7 + 4 *sqrt(3)) find irr(a,Q)
now let a= sqrt(7) + i find irr(a,Q)
$\displaystyle \sqrt{7 + 4\sqrt{3}}=2 + \sqrt{3}.$
$\displaystyle a^4 - 12a^2 + 64=0.$ show that the polynomial $\displaystyle x^4-12x^2+64$ is irreducible over $\displaystyle \mathbb{Z},$ and therefore over $\displaystyle \mathbb{Q}.$ in order to show this you only need to prove that we cannot have
now let a= sqrt(7) + i find irr(a,Q)
$\displaystyle x^4-12x^2+64=(x^2+ax+b)(x^2+cx+d),$ for some integers $\displaystyle a,b,c,d.$ that is very easy to verify!